Figure shows two identical parallel plate capacitors connected to a battery with the switch $S$ closed. The switch is now opened and the free space between the plates of the capacitors is filled with a dielectric of dielectric constant (or relative permittivity) $3$. Find the ratio of the total electrostatic energy stored in both capacitors before and after the introduction of the dielectric.

With the switch $S$ closed, the potential difference across capacitors $A$ and $B$ is same. So,

$V=\frac{Q_A}{C}=\frac{Q_B}{C}$

The initial charges on the capacitors are given by

$Q_A=Q_B=CV$

When dielectric is introduced, the new capacitance of either capacitor.

$C\text{'}=KC=3C$

Now, when the switch $S$ is opened, let the potential difference across capacitor $A$ be $V$ volt and that across the capacitor $B$ be $V\text{'}$ volt.

When dielectric is introduced with the switch open (i.e., battery disconnected) then, the charge on capacitor $B$ remains unchanged, so

$$\begin{gathered} Q_B=CV=C\text{'}V\text{'} \\ \Rightarrow V\text{'}=\frac{C}{C\text{'}}V=\frac{V}{3}volt \end{gathered}$$

Initial energy of both capacitors

$U_i=\frac{1}{2}C{V}^2+\frac{1}{2}C{V}^2=C{V}^2$

Final energy of both capacitors

$$\begin{gathered} U_f=\frac{1}{2}C\text{'}{V}^2+\frac{1}{2}C\text{'}{V}^2 \\ \Rightarrow U_f=\frac{1}{2}\left(3C\right)V^2+\frac{1}{2}\left(3C\right){\left(\frac{V}{3}\right)}^2=\frac{5}{3}C{V}^2 \\ \Rightarrow \frac{U_i}{U_f}=\frac{C{V}^2}{{\displaystyle \frac{5}{3}}C{V}^2}=\frac{3}{5} \end{gathered}$$